Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Compilation of tips and tricks to deal with remainders. [#permalink]

Show Tags

08 May 2011, 10:31

russty wrote:

hi thanks a lot for these tips on remainders. Kudos to u. I like the way u easily solved the 2 Remainder questions in some other posts. Could you also provide the links where you have solved questions involving the cycle of powers so as to get a better idea of that too.

thankssss

Hi,

Here's a link regarding remainders and cycles of power. I found it very helpful. I guess this link was mentioned somewhere in this forum only.I had stored it in my favorite list..hope u too find it useful.

Re: Compilation of tips and tricks to deal with remainders. [#permalink]

Show Tags

09 May 2011, 21:55

[/quote]

Hi,

Here's a link regarding remainders and cycles of power. I found it very helpful. I guess this link was mentioned somewhere in this forum only.I had stored it in my favorite list..hope u too find it useful.

Re: Compilation of tips and tricks to deal with remainders. [#permalink]

Show Tags

13 May 2011, 13:48

Just to clarify, for tip #5, if the numerator is smaller than the denominator then the remainder is simply the numerator? For example, 20/25 R = 20 and 4/13 R = 4?

Re: Compilation of tips and tricks to deal with remainders. [#permalink]

Show Tags

13 May 2011, 14:33

@seoulite Just to clarify, for tip #5, if the numerator is smaller than the denominator then the remainder is simply the numerator? For example, 20/25 R = 20 and 4/13 R = 4?

yes... coz, when you are dividing 4 by 13, quotient will be '0' and reminder will be 4.

Re: Compilation of tips and tricks to deal with remainders. [#permalink]

Show Tags

21 Feb 2012, 09:24

sriharimurthy wrote:

3) If a number leaves a remainder ‘r’ (the number is the divisor), all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor. Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 7 (which is a factor of 21) will also be 5.

If the value of ‘r’ is greater than the value of the factor, then we have to take the remainder of ‘r’ divided by the factor to get the remainder. Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 3 (which is a factor of 21) will be remainder of 5/3, which is 2.

ummm.....this cant be true.....say 147/35....remainder is 1.....7 is a factor of 35....so it should also leave remainder 1 ....but 147 /7 leaves a remainder 0..... am I missing smting?

Re: Compilation of tips and tricks to deal with remainders. [#permalink]

Show Tags

04 May 2012, 06:18

sriharimurthy wrote:

Hi guys, This is in conjunction with another post which has questions dealing with remainders (collection-of-remainder-problems-in-gmat-74776.html). I'm just trying to put together a list of tips and tricks which we can use to solve these kind of problems with greater accuracy and speed. Please feel free to comment and make suggestions. Hopefully we can add onto this list and cover all sorts of strategies that would help us deal with remainders! Cheers.

Please read this first : 1) Take your time with these points. Some of them might be a little difficult to follow in the first reading, but don't give up. The concepts are fairly simple. 2) These tips if mastered will be extremely valuable in the GMAT to help solve a variety of questions not limited specifically to remainders. I have been using them for quite a while now and they have not only helped me improve my accuracy but also my speed. 3) If you have any doubts, please do not hesitate to ask (no matter how stupid you might think them to be!). If you do not ask, you will never learn. 4) Lastly, have fun while trying to understand these tips and tricks as that, according to me, is the best possible way to learn.

All the best!

-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x-x- NOTE: Where ever you see R of 'x' it just stands for Remainder of x.

1) The possible remainders when a number is divided by ‘n’ can range from 0 to (n-1). Eg. If n=10, possible remainders are 0,1,2,3,4,5,6,7,8 and 9.

2) If a number is divided by 10, its remainder is the last digit of that number. If it is divided by 100 then the remainder is the last two digits and so on. This is good for questions such as : ' What is the last digit of.....' or ' What are the last two digits of.....' .

3) If a number leaves a remainder ‘r’ (the number is the divisor), all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor. Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 7 (which is a factor of 21) will also be 5.

If the value of ‘r’ is greater than the value of the factor, then we have to take the remainder of ‘r’ divided by the factor to get the remainder. Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 3 (which is a factor of 21) will be remainder of 5/3, which is 2.

4) Cycle of powers : This is used to find the remainder of \(n^x\), when divided by 10, as it helps us in figuring out the last digit of \(n^x\).

The cycle of powers for numbers from 2 to 10 is given below:

2: 2, 4, 8, 6 → all \(2^{4x}\) will have the same last digit.

3: 3, 9, 7, 1 → all \(3^{4x}\) will have the same last digit.

4: 4, 6 → all \(4^{2x}\) will have the same last digit.

5: 5 → all \(5^x\) will have the same last digit.

6: 6 → all \(6^x\) will have the same last digit.

7: 7, 9, 3, 1 → all \(7^{4x}\) will have the same last digit.

8: 8, 4, 2, 6 → all \(8^{4x}\) will have the same last digit.

9: 9, 1 → all \(9^{2x}\) will have the same last digit.

10: 0 → all \(10^x\) will have the same last digit.

5) Many seemingly difficult remainder problems can be simplified using the following formula : \(R of \frac{x*y}{n} = R of \frac{(R of \frac{x}{n})*(R of \frac{y}{n})}{n}\)

Eg. \(R of \frac{20*27}{25} = R of \frac{(R of \frac{20}{25})*(R of \frac{27}{25})}{25} = R of \frac{(20)*(2)}{25} = R of \frac{40}{25} = 15\)

Eg. \(R of \frac{225}{13} = R of \frac{(15)*(15)}{13} = R of {(2)*(2)}{13} = R of \frac{4}{13} = 4\)

6) \(R of \frac{x*y}{n}\) , can sometimes be easier calculated if we take it as \(R of \frac{(R of \frac{(x-n)}{n})*(R of \frac{(y-n)}{n})}{n}\) Especially when x and y are both just slightly less than n. This can be easier understood with an example:

Eg. \(R of \frac{(19)*(21)}{25} = R of \frac{(-6)*(-4)}{25} = 24\)

NOTE: Incase the answer comes negative, (if x is less than n but y is greater than n) then we have to simply add the remainder to n.

Eg. \(R of \frac{(23)*(27)}{25} = R of \frac{(-2)*(2)}{25} = -4.\) Now, since it is negative, we have to add it to 25.\(R = 25 + (-4) = 21\)

[Note: Go here to practice two good problems where you can use some of these concepts explained : numbers-86325.html]

7) If you take the decimal portion of the resulting number when you divide by "n", and multiply it to "n", you will get the remainder. [Special thanks to h2polo for this one]

Note: Converse is also true. If you take the remainder of a number when divided by 'n', and divide it by 'n', it will give us the remainder in decimal format.

Eg. \(\frac{8}{5} = 1.6\)

In this case, \(0.6 * 5 = 3\)

Therefore, the remainder is \(3\).

This is important to understand for problems like the one below:

If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t? (A) 2 (B) 4 (C) 8 (D) 20 (E) 45

hello thanks so MUCH for sharing please can anyone explain me the folloxwing formula or give example or actual gmat question where we can apply that

the formula is number 5 and 6 From your post

here 5) Many seemingly difficult remainder problems can be simplified using the following formula : \(R of \frac{x*y}{n} = R of \frac{(R of \frac{x}{n})*(R of \frac{y}{n})}{n}\)

6) \(R of \frac{x*y}{n}\) , can sometimes be easier calculated if we take it as \(R of \frac{(R of \frac{(x-n)}{n})*(R of \frac{(y-n)}{n})}{n}\)

Practice sri's wonderful tips and tricks: 4GMAT Home >> GMAT Test Prep Questions >> Number Systems...

(Can't post hyperlinks yet, sorry!)

I was wondering if someone could show me how to do this problem (found at the said site) quicker than my method (detailed below):

8. What is the remainder when the product of 1044, 1047, 1050, and 1053 is divided by 33?

I used sri's trick and found that 1056 is a multiple of 33. This resulted in remainders of (-12)(-9)(-6)(-3), respectively. Multiplied together, you get 1944. The remainder of 1944, when divided by 33, is 30, the correct answer.

Is there a quicker way than multiplying (-12)(-9)(-6)(-3) out?

Thanks, in advance!

Hi,

Im glad you found these tips helpful. There is in fact a quicker way to solve it.

R of \(\frac{(-12)*(-3)*(-9)*(-6)}{33}\) = R of \(\frac{(36)*(54)}{33}\) = R of \(\frac{(3)*(21)}{33}\) = R of \(\frac{63}{33}\) = \(30\)

As you can see, you don't really need to do any complex multiplications. Just multiply numbers is groups that yield a number closest to the denominator. That way you can keep simplifying to smaller numbers and avoid big calculations.

Let me know if anything needs to be clarified.

Cheers.

Hi,

What is the trick to find 1056 is a multiple 33? I tried this path but couldn't find the thread: GMAT Home >> GMAT Test Prep Questions >> Number Systems... Thanks,

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

I barely remember taking decent rest in the last 60 hours. It’s been relentless with submissions, birthday celebration, exams, vacating the flat, meeting people before leaving and of...

Rishabh from Gyan one services, India had a one to one interview with me where I shared my experience at IMD till now. http://www.gyanone.com/blog/life-at-imd-interview-with-imd-mba/ ...